Free surface boundary condition I am trying to verify the following identity. Considering the tensile component in the tangential direction on the free surface at a fluid-air interface
$$
\vec{s} \ \tau \ \vec{n} = 0
$$
where $\vec{s}$ is a tangent vector to the surface and $\vec{n}$ is the normal vector to the free surface, while $\tau$ is the stress.
Why is the above expression true? When I calculate the expression using
$$
\vec{n} = \frac{1}{\sqrt{1+h_x^2}} \left(-h_x,1\right)\,, \\
\vec{s} = \frac{1}{\sqrt{1+h_x^2}} \left(1,h_x \right)
$$
where $h_x = \partial h/\partial x$ and $h = h(x,t)$ indicates the depth from the bottom of the container to the free surface then
$$
\vec{s} \ \tau \ \vec{n} = \mu \left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right)\,,
$$
which is obviously not zero.
 A: Short answer. The condition you're looking for comes from the boundary condition (jump condition) at the interface between water and air, neglecting the viscosity of air.
Boundary/jump condition at interface.
Stress boundary condition at a surface comes from jump condition on stress, derived from the integral equilibrium of an elementary fluid volume, "with the thickness much smaller than the dimensions in the local plane of the surface", and reads
$\mathbf{t_n}_1 = \mathbf{t_n}_2$.
For a pair of fluids for which the stress vector can be written as the sum of pressure and viscosity contributions
$-p_1 \mathbf{\hat{n}} + \mathbf{s_n}_1 = 
-p_2 \mathbf{\hat{n}} +\mathbf{s_n}_2$,
being the viscosity contribution given as $\mathbf{s}_n = \mathbf{\hat{n}} \cdot \mathbb{S}$, where the viscous stress tensor can be written as $\mathbb{S} = 2 \mu \mathbb{D} + \lambda (\nabla \cdot \mathbf{u}) \mathbb{I}$ for a Newtonian fluid.
Boundary/jump condition at interface: components.
Now, we can write the jump condition separating the contributions in normal direction $\mathbf{\hat{n}}$ and tangential direction $\mathbf{\hat{t}}$, namely
$\mathbf{\hat{n}}: \qquad -p_1 + \mathbf{s_n}_1 \cdot \mathbf{\hat{n}} = 
-p_2  +  \mathbf{s_n}_2  \cdot \mathbf{\hat{n}} $
$\mathbf{\hat{t}}: \qquad \mathbf{s_n}_1 \cdot \mathbf{\hat{t}} = 
\mathbf{s_n}_2  \cdot \mathbf{\hat{t}} $
and these can be re-written as
$\mathbf{\hat{n}}: \qquad$$ [ -p_1 \mathbf{\hat{n}} + \mathbf{\hat{n}} \cdot (2 \mu_1 \mathbb{D}_1 + \lambda_1 (\nabla \cdot \mathbf{u}_1) \mathbb{I} ) ] \cdot \mathbf{\hat{n}} = [ -p_2 \mathbf{\hat{n}} + \mathbf{\hat{n}} \cdot (2 \mu_2 \mathbb{D}_2 + \lambda_2 (\nabla \cdot \mathbf{u}_2) \mathbb{I} ) ] \cdot \mathbf{\hat{n}}$
$\qquad \qquad\rightarrow \quad  -p_1 + 2 \mu_1 \mathbf{\hat{n}} \cdot  \mathbb{D}_1 \cdot \mathbf{\hat{n}} + \lambda_1 (\nabla \cdot \mathbf{u}_1)  =-p_2 + 2 \mu_2 \mathbf{\hat{n}} \cdot  \mathbb{D}_2 \cdot \mathbf{\hat{n}} + \lambda_2 (\nabla \cdot \mathbf{u}_2)$
$\mathbf{\hat{t}}: \qquad$$ \mathbf{\hat{n}} \cdot (2 \mu_1 \mathbb{D}_1 + \lambda_1 (\nabla \cdot \mathbf{u}_1) \mathbb{I} ) \cdot \mathbf{\hat{t}} = \mathbf{\hat{n}} \cdot (2 \mu_2 \mathbb{D}_2 + \lambda_2 (\nabla \cdot \mathbf{u}_2) \mathbb{I} ) \cdot \mathbf{\hat{t}}$
$\qquad \qquad\rightarrow \quad  2 \mu_1\mathbf{\hat{n}} \cdot  \mathbb{D}_1 \cdot \mathbf{\hat{t}} = 2 \mu_2\mathbf{\hat{n}} \cdot  \mathbb{D}_2 \cdot \mathbf{\hat{t}}$.
Boundary/jump condition at interface, in the limit of vanishing viscoisity of air.
Now, using the mapping 1:water, 2:air, and neglecting the viscosity of air $\mu_2 \ll \mu_1 \rightarrow \mu_2 = 0$, and neglecting volume viscosity of both fluids $\lambda_1 \approx \lambda_2 \approx 0$, we get
$\mathbf{\hat{n}}: \qquad p_1  = p_2$
$\mathbf{\hat{t}}: \qquad \mathbf{s_n}_1 \cdot \mathbf{\hat{t}} = 0 \qquad \rightarrow \qquad  2 \mu_1\mathbf{\hat{n}} \cdot  \mathbb{D}_1 \cdot \mathbf{\hat{t}} = 0 $
The last condition, if the interface is flat and we can use a set of Cartesian coordinates ($x$, $y$ in the plane of the surface, $z$ normal to it), reads
$x: \mu_1 \left[ \dfrac{\partial u_{x,1}}{\partial z} + \dfrac{\partial u_{z,1}}{\partial x}  \right] = 0$
$y: \mu_1 \left[ \dfrac{\partial u_{y,1}}{\partial z} + \dfrac{\partial u_{z,1}}{\partial y}  \right] = 0$
